If one root of the quadratic equation $i{x}^2-2\left(i+1\right)x+\left(2-i\right)=0, i=\sqrt{-1}$ is $2-i$, the other root is

Given quadratic equation is

 $i{x}^2-2\left(i+1\right)x+\left(2-i\right)=0$    Let $\left(i=\sqrt{-1} \Rightarrow i^2=-1\right)$

And, one given root is $\left(2-i\right)$.

We have to find the other root.

Using the fact of quadratic equation $a{x}^2+bx+c=0$ that

Sum of roots $=\frac{-b}{a}$

and product of roots $=\frac{c}{a}$

So, for the given quadratic equation,

$$\begin{gathered} \text{Sum of roots}=\frac{-\left(-2\left(i+1\right)\right)}{i} \left[\begin{array}{c}\because b=-2\left(i+1\right)\\ a=i\end{array}\right] \\ =\frac{2\left(i+1\right)}{i}=\frac{2\left(i+1\right)}{i}\times \frac{i}{i} \left[\text{Rationalising}\right] \\ =\frac{2\left(i^2+i\right)}{i^2}=\frac{2\left(-1+i\right)}{-i} \left[i^2=-1\right] \\ =2\left(1-i\right)=\left(2-2i\right) \end{gathered}$$

Given one root is $\left(2-i\right)$

So, the other root of the given quadratic equation is sum of roots minus one root value.

$$\begin{gathered} \therefore \text{Other root}=\left(\text{Sum of roots}\right)-\left(\text{Given one root}\right) \\ =\left(2-2i\right)-\left(2-i\right) \\ =\left(2-2i-2+i\right)=-i \end{gathered}$$

So, the other root is $-i$.

Hence, option $\left(1\right)$ is correct.